1
Fundamentals of Engineering Review for Dynamics
Dr.’s Yannitell (retired)
&
Dr Waggenspack [email protected]
Louisiana State University
__________________________________________________________ Last Revised: 02/14
Review Notes available in PDF format @ http://www.eng.lsu.edu/students/current/feinfo.html
Dynamics Problem Decomposition Dynamics!
Kinetics!Kinematics!
m
!=5°
Pm
Lµ != 0.4k
A
!r
30°
300 mm
O
v!!, aA A
Geometric descriptions of motion & constraints
Loading relationships which dictate CHANGES in motion
2
Dynamic Studies Dynamics:!
(Kinematics & Kinetics)!
Particles!
1 mh
v02.8 m
!20 m
• M - mass • (x,y) - position in 2D • 2 Degrees of Freedom
(DOF) in the plane
• M & I - both mass & rotational inertia
• (x,y,θ) - position & orientation in 2D (3DOF)
Rigid Bodies!
4 cm
!
A = (x,y)
B
x
y
Getting Started => Particle Kinematics • Rectilinear Motion
– Movement along a straight line in 1-2 or 3D • 1 Degree of Freedom (DOF)* - s(t)
X
Y
O
e
s
r
p
+s
-s
• Curvilinear Motion – Movement of particle along an arbitrary path
through space
p
isi
s=0s(t)
X
Y
O
r i
3
Rectilinear Motion Overview (Calculus/Physics Review!):
• Position - s(t)
(1) v = dsdt
= s•
• Speed - v(t)
• Acceleration - a(t)
(2) a = dvdt
= v•=d2sdt 2 = s
••
DIFFERENTIATE
INTEGRATE
+s
-s
ts
+v
-v
v
+a
-a a
• Typical Functions ?? – Polynomials – Trigonometric – Logarithms – Exponentials
s min
v = 0
v min
a = 0
dsdt
dv dt
• Basic Calculus! – Max’s & Min’s ? – Undo Differentiation?
Rectilinear Motion Summary: • Position - s(t)
(1) v = dsdt
= s•
• Speed - v(t)
• Acceleration - a(t)
(2) a = dvdt
= v•=d2sdt 2 = s
••(2*) v dv = a ds
• Alternate form ?
DIFFERENTIATE
INTEGRATE
a(t) => Solid Rocket Propulsion a(v) => aerodynamic drag
a(s) => Gravitational fields, springs, conservative forces etc.
+s
-s
ts
+v
-v
v
+a
-a a s min
v = 0
v min
a = 0
4
Given: s =2t2 - 8t + 3 Find: Displacement from t = 1 to t = 3 Distance traveled from t = 1 to t = 3
Since, s(3) - s(1) = 0 displacement = 0
distance traveled = | s(2) - s(1) | + |s(3) - s(2) | = | -5 - (-3) | + | (-3) - (-5) | = 2 + 2 = 4
-8
3
-3 -5 -3
4
s
v t
t
1 2 3
s = 2t2 - 8t + 3 v = s = 4t - 8 a = s = 4 ••
• s(2) = -5 v(2) = 0 a(2) = 4
s(1) = -3 & s(3) = -3
Reversal @ v|t=2 = 0
Rectilinear Kinematics: Accel. a function of velocity – a(v)
t f = t(v) =dva(v)v i
v f
! " ti =dv"kv28
v f
! + 0
Given: • A freighter moving at 8 knots when engines are stopped • Deceleration a= -kv2
• Speed reduces to 4 knots after ten minutes
! t f =1kv 8
v f
=1k1v f"18
#
$ % %
&
' ( ( ! vf = v(tf ) = 8
8kt f + 1 (knots)
v = 8 knots a = - k v 2
s0
Solution: (A) With a, v & t parameters given/requested, use a=dv/dt form
Find: (A) Speed of the ship as a function of time v(t) (B) How far does the ship travel in the 10 minutes it takes to
reduce the speed by 1/2 ?
5
and the resulting expression for speed of the ship as a function of time v(t) is as follows
v f = v(t) =8
6t +1
(knots)
• Substituting BCʼs helps resolve the unknown constant k!
t =10 (min)
60 (min/hr)=
16
hr , v = 4 knots
! k =
34 1
nm"
# $ $
%
& ' ' ! v(1/ 6) = 8
8k 1/ 6( ) +1= 4 (knots)
• From here, there are two alternatives for resolving the second question!
v = 8 knots a = - k v 2
s0
Rectilinear Kinematics: Accel. a function of velocity – a(v)
and the resulting expression for position of the ship as a function of time s(t)
(B) !METHOD 1: Now, knowing the velocity as a function of time!
v(t) = 86t +1
=dsdt
ds0
sf
! =8
6t +1
dt0
t f
!
s f ! 0 = 43
ln 6t +1( ) 0
t f =43
ln 6t +1( ) ! ln(1)( )
s f = s(t) =4
3
ln 6t +1( )
can now be used to find the particular displacement/distance at t=1/6 hr !
s(1/ 6) = 4 3
ln 6(1/ 6) +1( ) =4
3
ln(2) (nautical miles)
v = 8 knots a = - v 2
s0
34
!the boatʼs position can be found by integration!
Rectilinear Kinematics: Accel. a function of velocity – a(v)
6
and as was seen before
(B) !METHOD 2: With a, v & s parameters given/requested, use ads=vdv form!
s(t = 1/ 6)! s(v = 4) = 4 3
ln(2) (nautical miles)
s f = s(v) =vdva(v)vi
v f
! + si " s(4) = vdv#3/ 4v2( )8
4
! + 0
s(4) = !4
3dvv8
4" = !43 ln v 8
4=!43 ln4! ln8( ) = 4
3ln 84
Q.E.D.
v = 8 knots a = - v 2
s0
34
!and the boatʼs displacement (position?) can again be found by integration!
Rectilinear Kinematics: Accel. a function of velocity – a(v)
Curvilinear Kinematics Summary: • Position
r(t)= x(t)i + y(t)j+ z(t)k= r(t)er + zez
• Velocity: v = d r / dt
v(t)= r•
(t)= x•
i+ y•
j + z•
k
= ve t = s•
e t = r•
er + r!•
e! + z•
ez
• Acceleration: a = d v / dt
a(t)= v•
(t)= r••
(t) = x••
i + y••
j + z••
k = v•
e t + !"• 2
en = v•
e t +v2
!en
= r••
# r"• 2$
%
& & &
'
(
) ) ) er + r"
••
+ 2r•
"•$
% & &
'
( ) ) e" + z
••
ez
O
p
r(t)
v(t )
a(t )
7
Curvilinear Kinematics Summary: • Position
r(t)= x(t)i + y(t)j+ z(t)k= r(t)er + zez
• Velocity: v = d r / dt
v(t)= r•
(t)= x•
i+ y•
j + z•
k
= ve t = s•
e t = r•
er + r!•
e! + z•
ez
• Acceleration: a = d v / dt
a(t)= v•
(t)= r••
(t) = x••
i + y••
j + z••
k = v•
e t + !"• 2
en = v•
e t +v2
!en
= r••
# r"• 2$
%
& & &
'
(
) ) ) er + r"
••
+ 2r•
"•$
% & &
'
( ) ) e" + z
••
ez
er
O
e!
ref. line
r(t) p2D
!
v(t )
i
j
ere!
et
en
!
"a
2D Curvilinear Motion: Coordinates & Conversions
• Cartesian <-> Polar <-> Path r(t) = xi + y j = rer
er = cos !i + sin!j = x
ri+ y
rj
e! = k " er = cos ! j # sin!i
i = cos!er " sin!e! j = k ! i = cos"e" + sin"er
v( t) = r•(t) = x
•i+ y
•j = ve t = s
•e t
et =vv=x•
vi+ y
•
vj
en = k ! e t =x•
vj " y
•
vi
i = cos!et " sin!enj = k ! i = cos"en + sin"et
er
O
e!
ref. line
r(t) p2D
!
v(t )
i
j
ere!
et
en
!
"a
8
15
Curvilinear Motion: Cartesian Coordinates!
• Projectile Motion
a( t) = 0i ! g j = 0, !g[ ]• z component drops out!
– Scale w.r.t. earth such that gravity g is ~constant |g| = 32.2 ft/s2 = 9.81 m/s2
– Neglect any air resistance – Motion is PARABOLIC thus PLANAR! – Typically align
• y-axis along gravity vector • x-axis horizontal in direction of motion
a = g
O x
y
r=(x,y)
v
16
Curvilinear Motion: Projectile Motion!
– Integrate rectilinear relations • Two (2) scalar relations • One VECTOR relationship a(t) = 0i! g j= 0,!g[ ] = ac
! dvvi
v f" = ac dtti
t f"
v f ! vi( ) = ac t f ! ti( ) ! drri
r f" = v(t)dtti
t f"! v f (t) = ac t f " ti( )+ vi ! r f =
ac2t f " ti( )
2+ vi t f " ti( )+ ri
vx f = 0 t f ! ti( )+ vxi = vxivyf = -g t f ! ti( )+ vyi
x f = vxi t f ! ti( ) + xiyf =
-g2tf ! ti( )
2+ vyi tf ! ti( ) + yi
O x
y v = ( v ,v )
a = g = (0,-g )r =(x ,y )i i i
i x yii
9
17
Curvilinear Kinematics: Projectile Motion example ref: Hibbler 12-104!
Given:!– Figure shown w/ ground !– t0=0, (x0,y0)=0, v0=v0 @ θ above horizon!
y = !kx 2
Find: In terms of v0 , θ & k – (A) The location at impact (xI,yI) – (B) Velocity & Speed @ impact, vI, vI – (C) Elapsed time @ impact, tI
Solution: – 2D projectile motion – Get expressions for vx(t),vy(t) then x(t), y(t) – Substitute into ground constraint expression
• Solve for time of impact – With tI known, substitute & solve for (xI,yI)
y = - kx 2
O x
y a = g = (0,- g )
v = v @ θ 0 θ
(x ,y )I I
! v f (t) = ac t f " ti( )+ vi ! r f =ac2t f " ti( )
2+ vi t f " ti( )+ ri
18
Curvilinear Kinematics: Projectile Motion!
a( t) = 0i ! g j = 0, !g[ ]
! (B) v I = a tI " 0( ) + v0 = vx I ,vyI[ ] vx I
= v0 cos!
vy I= -gtI + v0 sin!
• IC’s => t0=0, (x0,y0)=0, v0=v0 @ θ!
s•= v = vxI
2 + vy I2
= v0 cos!( )2 + -gtI + v0 sin!( )2
= v02 " 2gv0 sin!tI + gtI( )2
– Speed y = !kx 2
O x
y a = g = (0,-g )
v = v @ "0
(x ,y )I I
!
10
19
Curvilinear Kinematics: Projectile Motion!
! (A) rI =a2tI " 0( )2
+ v0 tI " 0( ) + 0
xI = v0cos! tI
yI =-g2tI
2 + v0sin! tIy = !kx 2
O x
y a = g = (0,-g )
v = v @ "0
(x ,y )I I
-g2tI
2 + v0sin! tI = "k v0cos! tI( )2
! (C) tI =2v0sin"
g # 2k v0cos"( )2
, tI = 0
! (B) v I = a tI " 0( ) + v0 = vx I ,vyI[ ]
– Substitute value for tI into position, velocity & speed relations for solution
y = !kx 2
20
Curvilinear Motion: Projectile Motion!
a( t) = 0i ! g j = 0, !g[ ]
– Other typical P.M. queries • Max Height • Max Range • Time @ some place along trajectory • Later w/ Path & Polar Coord
– Velocity (speed,direction/tangent) – Curvature, rate of speed change ….
! v f = a tf " ti( ) + vi ! r f =a2tf " ti( )2 + v i t f " ti( ) + r i
O x
y v = ( v ,v )
a = g = (0,-g )r =(x ,y )i i i
i x yii
– Reconsider problems w/ different axes placement
11
21
Given: launch at 3600 m altitude vo = 180 m/s angle 30°
x = 0 y = -9.81
for h set y = 0 t = 9.17 h = y = 4013 m
x = 180 (cos 30) = 156 y = 180 (sin 30)-9.81 t = 90-9.81 t x = 156 t y = 90 t - 4.905 t2 +3600
for tT set y = 0 t = 31.18
.. ..
. .
3600 h
30° 180 m/s
T
.
a = - g j
Path Coord. Example ref: Meriam&Kraige 2-8
Given: – A rocket at high altitude with – a0=6i-9j (m/s2) – v0=20 (km/hr) @ 15° below horizontal
Find: At instant given (A) The normal & tangential
accelerations (B) Rate at which speed is increasing (C) Radius of curvature of the path (D) Angular rotation rate of the radial from CG to center of curvature
cg
C
Solution: – “High altitude” means negligible air resistance – Interested only at this instant (NO Integration required) – V given is TANGENT TO THE PATH
• Use this to relate path to cartesian coordinates
v = 15 km/hr
O x
y15°
a = g = (6,-9)20
12
Path Coord. Example ref: Meriam&Kraige 2-8
Solution (cont’d):
!
e t =vv
=vv
= cos15°i " sin15°j
!
en = ± k " e t( )= # cos15°j# sin15°i
(A) an & at =?
(C) ρ=?
at = a • e t = 6i ! 9j( ) • cos15°i! sin15°j( ) = 8.12 (m / s2 ) = at
!
an = a •en = 6i " 9j( ) • "cos15°j" sin15°i( ) = 7.14 (m /s2) = an
v•= at = 8.12 (m / s2 )(B) v
•= ??
an =v2
! " ! =
v2
an
=(20 *103km /hr )2
7.14(m / s2 )1hr
3600s* 103m
km#
$ % %
&
' ( (
2
= 4.32(106 )m
v = 15 km/hr
O x
y15°
a = g = (6,-9)
cg
C
eten 20 (2D shortcut!)
Path Coord. Example ref: Meriam&Kraige 2-8
Solution (cont’d):
an = !"• 2
#"•=
an!=
7.14 (m / s 2 )4.32 *106(m)
= 12.9 *10$4 1s
(D) – Look either at an or velocity !•= ??
v
= !"•
# "• =
v!
=20*103km / hr4.32 *106 (m)
1hr3600s
* 103mkm
$
% & &
'
( ) ) = 12.9 *10*4m
v = 15 km/hr
O x
y15°
a = g = (6,-9)
cg
C
eten 20
13
RELATIVE MOTION
rB = rA + rB/A vB = vA + vB/A aB = aA +aB/A
When A & B are two points on the same rigid body then: • i.e. the relative motion is circular • vB/A is perpendicular (⊥) to rB/A & • | vB/A | = | ωAB AB |
rB
B
rB/A
A rA
rB/A
B
A vB/A
Special Case: Rigid Bodies
vB = vA + ωAB k × ABuB/A (2D)
ωAB
26
Relative Motion: ref ~DWY FE Review
Given:!• A river flows south at 5 m/s!• A boat can travel at 10 m/s relative
to the water.!Find: • In what direction should the boat
head from A, in order to reach point B, directly across the river?
Solution: • Set up Cartesian Axes @A • Write Relative Velocity expression
VB = VR + VB/R DM DM DM
A B
v = 5 m/sR
x
y
Ov
v = -5 j m/sR
v = ? i B
v = 10 m/s @! ?B/R
!• Draw Velocity Polygon diagram
– VR magnitude & direction known – VB direction known, not magnitude – VB/R magnitude known, not direction
14
27
Relative Motion: ref ~DWY/WNW FE Review
A B
v = 5 m/sR
x
ySolution (cont’d):
VB = VR + VB/R DM DM DM
vB i = vR j + vB/R (cos θ i + sin θ j) vB i = -5 j + 10 (cos θ i + sin θ j) (m/s)
• Equating components: i => vB = 10 cos θ (m/s) j => 0 = -5 + 10 sin θ (m/s)
• 2 equations <-> two unknowns v = ? i BOv
v = -5 j m/sR
v = 10 m/s @!B/R
!
sin θ = 5/10 = 1/2 (m/s) => θ = sin-1 (1/2) = 30°
• Lagniappe vB = 10 cos 30° (m/s)
28
Relative Motion: ref ~DWY/WNW FE Review
Alternate Solution (cont’d):
A B
v = 5 m/sR
x
y
• With velocity polygon drawn, use the Law of Sines
v = ? i BOv
v = -5 j m/sR
v = 10 m/s @!B/R
!
sin!vR
=sin90o
vB/R
! = sin"1 vR sin90o
vB/R
= sin"1 510
#
$ % %
&
' ( ( = 30o
θ
15
29
Relative Motion: ref ~Meriam & Kraige 2/13
Given:!• Two cars A & B at the instant shown!
vA= 72 i km/hr !aA= 1.2 i m/s2!
vB= 54 et km/hr, constant speed!
Find: (A) vB/A =? (B) aB/A =?
Solution: • Convert to consistent units
x
y
A
30°
B
150 m e n
e t
en = cos30°i ! sin30°je t = k ! en = cos30°j + sin 30°i
(km /hr)* 13.6
= (m / s) ! vA = 72(km /hr) = 20(m / s)vB = 54(km /hr) = 15(m / s)
• Motion RELATIVE TO A of interest • Two coordinate axes are used
– Simplifies v & a definitions – Illustrates “coordinate conversion” for
expressing answers “in terms of” a unified set.
30
Relative Motion: ref ~Meriam & Kraige 2/13 (A) !Relative Velocity!
x
y
A
30°
B
150 m e n
e t
vB/A = vB ! vA
=15e t ! 20 i (m / s)=15(sin30°i + cos30°j) ! 20 i (m / s)
vB/A = !12.5i +13.0j (m / s) = 18 (m / s)@! 46°
• Velocity Polygon Approach (Graphical)
v = 20 m/sA
v = 15 m/sB
Ov
30°
46°
vA
vB! vB/A = vB " vAvB = vA + vB/A
VB/A = 18 m/s!
16
31
Relative Motion: ref ~Meriam & Kraige 2/13 x
y
A
30°
B
150 m e n
e t
)/( 2.1)30sin30(cos5.1
)/( 2.1 5.12
2
smsm
ijiie
aaa
!°!°=
!=
!=
n
ABB/A
°!=!= 82@)/(76.0)/( 75.0 1.0 22 smsmjiaB/A
• Acceleration Polygon (Graphical)
(B) Relative Acceleration aA = 1.2 i (m / s2 )
aB = v• e t +
v2
! en (m / s 2) = 15 m / s( )2
150 men
= 1.5en (m / s2 )
a = 1.2 m/s2A
a = 1.5 m/sB2
Oa30°
a = 1.5 m/sB2
a = 1.2 m/s2A
aB = aA + aB/A! aB/A = aB " aA
-82°!aB/A = 0.76 m/s2!
32
Given: A balloon at an altitude of 60 m is rising at steady rate of 4.5 m/s. A car passes below at constant speed of 72 kph.
Find: Relative rate of separation 1 second later: aC = 0 aB = 0 (m/s2)
vC = 20i vB = 4.5j (m/s)
72 km/hr
60 m
4.5 m/s
Alternative Method (Vectors!):
!
r•B /C =V B /C •
rB /C
rB /C
= V B "VC( ) •rB " rC( )rB /C
= "20,4.5( ) •"20,64.5( )
67.5=
690.367.5
=10.2 (m/s)
( )222B/C 5.460)20(r tt ++=
5.4)5.460(2)20)(20(2r2r B/CB/C tt ++=•
Divide through by 2 rB/C & set t = 1 )/(22.1052.67/25.690r / smCB ==
•
rC = 20t i rB = (60 + 4.5t)j (m)
= 20 m/s
J
I
rB/C = rB - rC
j
i
rB/C = rB/C = rB/C •rB/C
17
33
Given: The ferris wheel rotates at θ= 2 r/s, θ = -1 r/s2 and the boy (B) walks to the right at a constant speed of 2 m/s.
Find: The velocity and acceleration of girl (G) on the ferris wheel relative to boy B
B
R= 4m θ
G
j i
eθ er
• •
Defining unit vectors i, j, er, & eθ shown velocities & accelerations are then:
vB = 2i (m/s) aB = 0
•
vG = 4 m (2 r/s) eθ = 8 eθ (m/s) aG = -4 m (2 r/s)2 er + 4 (-1 r/s2) eθ = -16 er - 4 eθ (m/s2)
Solution:
=> vG/B = vB - vG = ? aG/B = aB – aG = ?
θ •θ • •
er = cos θ i + sin θ j eθ = -sin θ i + cos θ j
or i = cos θ er - sin θ eθ j = sin θ er + cos θ eθ
vG/B = vG - vB = 8 eθ - 2i = -(8 sin θ + 2) i + 8 cos θ j OR = -2 cos θ er + (2 sin θ + 8) eθ (m/s)
aG/B = aG = -16 er -4 eθ (m/s2)
Must convert axes to combine vectors:
and since aB = 0 :
ωAB
VB
VA
B
vB = vA + vB/A
vB j = vA i + ωAB k × AB uB/A
vB j = vA i
- AB ωAB(sinθ i + cosθ j)
θ
A
Equating i & j components:
i → vA - AB ωAB sinθ = 0
j → vB = AB ωAB cosθ
vA AB ωAB sinθ vB AB ωAB cosθ =
Two points on a rigid body:
18
VA = AC ω AB [ i ]
VB = - BC ωAB [ j ] ωAB
C I.C.
VB
VA
B
θ
A
Using Instant Centers (IC):
AC = AB sinθ BC = AB cosθ
vA AB ωAB sinθ vB AB ωAB cosθ =
O
ωo
VB ⊥ OB A VA
B
VB
VA
VB/A ⊥ AB
Slider Crank Velocities Using Graphical & Instant Centers (IC): C
ωAB
VB = OB ωo = CB ωAB VA = CA ωAB Be sure to account for direction!
VA = (OB / CB) CA ωo
19
Given: ωc = 2 r/s αc = 6 r/s2
Find: vD, aD
AC
ωc αc
D
| 6”| 6"| 8” |
D
CA
B
F E2 E1
rA = 6" rB = 12" rC = 8"
ωC = 2 r/s αC = 6 r/s2
VE1 = rCωC = 8•2 = 16 in/s
VE2 = 16 = rBωB = 12 ωB
so: ωB = 4/3 = ωA [r/s]
VD = VF = ωArA = 4/3 (6) = 8 [in/s]
aE1 = αCrC + ωC2rC
= 6 (8) + 4(8) = 48 + 32
aE2 t = aE1t = 48 = αBrB = αB(12) αB = 4 = αA
aF t = αArA = 4(6) = 24 = aD
[in/s2 ]
[in/s2]
20
Given: ro = 3' ri = 2' vo = 10 f/s no slip
Find: vB
vo = 10 ft/s
vc = vo + ω r = 10 - 2 ω = 0
ω = 5 or -5 k
vB = vo + ω x rB/o = 10 i + -5k x -3j = -5 i [ft/s]
or vB = vc + ω x rB/c = 0 + 5k x -1j = -5 i [ft/s]
O 45°
A
C
B
vA = ?
Kinetics Summary • Three general solution approaches for establishing the governing
equations of motion (EOM) => Which one to use? i) Newton’s Laws ii) Work- Energy & Conservation of Energy iii) Impulse - Momentum & Conservation of Momentum
MP! = ICG " + reff maCG
UA!B = matdssA
s B
" = mvdvvA
vB
" = 12m vB
2 ! vA2( ) = #TA!B
– Typical forces • Springs • Friction • Gravitation
F f = µs / k NF = k s - s0( )
F =mg
UNC = !T + !Vg + !Ve = !ETOT
I = FRdt! = dL! = "L
F! = m aCG
21
Particle Kinetics: Free Body Diagrams!• Free Body Diagrams:!
– Isolate the particle/system of interest (i.e. boundaries)!– For noting action-reaction between particles/bodies it is
important to identify the common normal-tangent @ the point of contact (often one or the other is easily identified)!
mAm B
FAB
FBA
The image part with relationship ID rId6 was not found in the file.
– Include ALL forces (& later => moments)!• Field forces (gravity, electro-magnetic fields etc)!• Viscous forces (aerodynamic drag, fluid flows, etc)!• Contact forces (touching elements) -- Most common!
– For motion over an interval --- draw in a general position!!
n!
t!
n!
t!
Kinetics: Σ F = maC Σ MC = IC α
F2 C = cg
etc.
M1
F1
C IC α
maC
FBD EFD
MP! = IC " + reff maC
22
Given: the 20# force is applied to the sliding door which weighs 100 #
6'
10 '
5'
20#
Find: the reactions at the frictionless roller supports
Find: the reactions at the frictionless roller supports
ΣF → = 20 = (100/g) a a = g/5
ΣMA = 10B - 5(100) + (20) = 3 (100/g)(g/5)
=> B = 54 #
A = 46 #
6' 5'
20#
A B10 '
100# 100 g a
ΣF = A + B - 100 = 0 ΣMCG = 5B - 5A - (20) 2 = 0
23
α
O!45°!
#1!
#2!
100 mm!
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Given:!• The slider (m=2 kg) fits loosely in the smooth slot of
the disk which lies in a horizontal plane and rotates about a vertical axis through point O.!
• The slider is free to move only slightly along the slot in either direction before one (but not both) of the two wires #1 or #2 becomes taut. !
• The disk starts from rest at time t = 0 and has a constant clockwise angular acceleration of α=0.5 r/s2.!
Find:!(A) Determine the TENSION (T2) in wire #2 at t =1 second!(B) Determine the REACTION FORCE (N) between the
slot and the block, again at t =1 second.!(C) Determine the TIME (t ) at which the tension in wire #2
goes slack and wire #1 becomes taut. Solution:!• Asks for FORCES (T,N) so we must first establish kinematics (accelerations!)!• “Move only slightly” means it is effectively fixed relative to the slot/disk, thus!• The slider travels a circle about O & path (en-et) axes !! ! ! ! !or polar (er-eθ) axes are convenient!
er!
eθ!
er!
eθ!
en!
et!
en!et!
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!• Construct FBD!• Use disk kinematics (α=0.5 r/s2 CW constant) to
determine sliderʼs total acceleration!
α
O!45°!
#1!
#2!
100 mm!er!
eθ!
en!
et!
en ,-er!
et ,eθ
45°!
T2 or -T1 !
N!
!
" = 0.100m = r # constant
!
" #•
= #••
= r•
= r••
= 0
!
d"0
"
# = $dt0
t# = 0.5dt
0
t#
" = 0.5t
• Not instantaneous - integrate angular acceleration!
!
a s= " re# $ r% 2 er
= v•
e t +v 2
&en = " re t + % 2 ren
24
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!• Newtonʼs Law can be applied along ANY two
independent directions to resolve unknown reactions!– Sum force components along (n-t, r-θ)
α
O!45°!
#1!
#2!
100 mm!er!
eθ!
en!
et!
en ,-er!
et ,eθ
45°!
T2 or -T1 !
N!
!
T2 cos45+N sin 45 = m"rT2 sin 45#N cos45 = # m$ 2r
!
T = m " r cos45#$ 2r sin 45( ) =mr 2
2" #$ 2( )
N = m " r cos45+$ 2r sin 45( ) =mr 2
2" +$ 2( )
– OR to simplify algebra of unknowns, choose the directions along the unknown reactions and sum both forces and acceleration components!
– ASIDE: This IS the geometric equivalent to simultaneously solving the first set of constraints to yield expressions for the unknowns!
– Noting the similarity of the expressions (± : + for N, - for T)
!
N,T2 =mr 22
" ±# 2( )
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Solution (continued):!• Substituting the known expressions for α & ω(t) α
O!45°!
#1!
#2!
100 mm!er!
eθ!
en!
et!
en ,-er!
et ,eθ
45°!
T2 or -T1 !
N!
(A) !So for t=1, the TENSION T2 is !
N,T2 =mr 2
2" ±# 2( )
=2kg*0.1m* 2
2 0.5 ± 0.5 t( )2{ }(r /s2)
N,T2 =2
20 1 ± 0.5 t 2{ }(N)
!
T2 =2
20 1 " 0.5(1)2{ } (N )= 2
40(N ) = 0.035(N )
(B) !At t=1, the NORMAL REACTION N is
!
N =2
20 1 + 0.5(1)2{ } (N ) =
3 240
(N ) = 0.106(N )
(C) !The time when TENSION T2 goes to zero is!
!
T2 =2
20 1 " 0.5t2{ } (N ) = 0 #1 " 0.5t2 = 0 # t= 2 # t= 1.414 (s)
25
en ,-er!
et ,eθ
45°!
T2 or -T1 !N!
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/74!Langiappe:!• The acceleration vector starts off completely in the lateral (θ or t) direction here
(ω=0). Since cables/wires/ropes cannot PUSH, only T2 can be engaged in balancing the (r or n) component of the side wall reaction N!
• The tangential acceleration component remains constant!• As the disk speeds up (ω >0), the normal component increases!• When the total acceleration vector aligns with the normal reaction force
between the block & slot, the cord/wire tensions are both zero momentarily, and as T2 goes slack, T1will become taut.!
increasing ω
T1=T2=0!!
a t= 0 = a t or a"
Impulse / Momentum
If mutual forces cancel, impulses cancel ( not true of work )
coef. Of restitution e =
Angular Momentum:
rel. norm. sep. vel. rel. norm. app. vel.
∫ F dt = ∫ madt
= ∫ mdv
= mv2 - mv1
t2
t1
∫ Mc dt = ∫ Ic α dt
= ∫ Ic dω
= ∫ Ic ω2 - Ic ω1
° °
°
° °
26
Example: Conservation of Momentum Given: • An artillery gun (mG) fires a shell
(mP) with a speed vp Find:
(A) The recoil speed (vR) of the gun
Solution: • FBD of system components, just as
shell leaves the gun • Rectilinear motion (i.e. only horizontal
motion of interest here) • Propellant firing is internal to the
system – System momentum is conserved in the
horizontal direction
vR
vP
FBD! mGg! mPg!
Fpropellant!
N!
!
"Lsys = 0
x!
!
"Lx#system = mG (vG # 0) + mP (vP # 0) = 0
!
vR = " vG = mP
mG
vp
Example: Conservation of Momentum Given: • More often, a “muzzle velocity” (vP/G)or
speed of the shell relative to the gun barrel is specified
Find: (A) The recoil speed (vR) of the gun Solution: • FBD (same), Rectilinear motion & Propellant
firing is internal
FBD!
vR
vP/G
mGg! mPg!
Fpropellant!
N!
!
"Lsys = 0
x!
!
mGvG + mP (vG + vP /G ) = 0
!
vR = " vG = mP
mG + mP
#
$ %
&
' ( vP /G
!
vP = vG + vP /G
!
"Lx#system = mG (vG # 0) + mP (vP # 0) = 0
27
Example: Conservation of Momentum!Given:!• Numerous examples with similar
circumstances, rephrasing the wording!– Kid(s) on a boat in still water, one
jumps off!– Car lands on a barge & skids to rest
relative to barge!– Rail cars collide & stay attached!
Find:!(A) The resulting speeds of each element!(B) A time it takes to “skid to rest” !
Solution:!• Similar conservation of momentum relations !
!
"Lsys = 0
A B C
50 km/hr
A B
• Impact Problems: – Reformulation of one type of Impulse-Momentum
Particle Kinetics: Impulse-Momentum
– Impulsive Forces characterized by • LARGE MAGNITUDE • SHORT TIME DURATION • Ex: explosions, ball-bat, club-golf ball
– Neglect other conventional forces of lesser effect for the short time interval
• Springs • Gravity • Many Reaction forces (BUT NOT ALL!)
– Good opportunity to look at the SYSTEM of particles in simplifying the problem (reactions are internal!)
L = mv|F |
t (sec) Δt
n
t
F
F
28
Particle Kinetics: Impulse-Momentum/ Impact • Impact
– Locate Common Normal/Tangent • Line of contact/impact - the NORMAL!
– Forces of interaction • Equal, Opposite, Co-linear
– Very complex internal phenomena, captured by Coefficient of Restitution
(good derivation in B&J text --- READ IT!)
– Central & Oblique Impacts • Velocities are NOT co-linear with the line
of impact (i.e. the common normal)
n
t
F
F
n
t
vB
A
B
vA
vA*
Oblique* Impact
Central Impact
vB*
e =
Vrelative Separation( )Vrelative Approach( ) Common Normal
Particle Kinetics: Impulse-Momentum/ Impact • Solving Impact Problems !
n
t
vA
vB
vA*
vB*
A
B
(1) Tangential Direction: individual particles have no net external impulsive forces in!
mAvAt =mAvAt* & mBvBt = mBvBt
*
(2) System of particles: No net impulsive forces normal direction!
!Ln SYS= 0" mAvAn +mBvBn = mAvAn
* +mBvBn*
(3) Coefficient of Restitution: Rel. Velocities along Common NORMAL!
e = vRelative Separation
vRelative Approach Normal
=vBn* ! vAn
*vAn ! vBn
!
(Perfectly Plastic) 0 " e "1 (Perfectly Elastic)
29
Particle Kinetics: Impulse-Momentum/ Impact!• Solving constraint relations !!
n
t
vA
vB
vA*
vB*
A
B
*BtBt
*AtAt vvvv == &
( ) (1) *AnAnBn
*Bn vvvv !+=
B
A
mm
( ) **(2) AnBnAnBn vvvv +!= e
• From which the unknown rebound (normal) component of velocities become!
( ) 1 (1) BnAn*An vvv e
mmm
mmemm
BA
B
BA
BA +!!"
#$$%
&
++!!
"
#$$%
&
+'
=
( ) 1 (2) BnAn*Bn vvv !!
"
#$$%
&
+'
++!!"
#$$%
&
+=
BA
AB
BA
A
mmemme
mmm
Particle Kinetics: Impulse-Momentum/ Impact!• What if mA >>> mB ?!
n
t
vA
vB
vA*
vB*
A
B
*BtBt
*AtAt vvvv == &
( ) (1) *AnAnBn
*Bn vvvv !+=
B
A
mm
( ) **(2) AnBnAnBn vvvv +!= e
• From which the unknown rebound (normal) component of velocities become!
( ) 1 (1) BnAn*An vvv ee ++!=
(2) Bn*Bn vv =
0
30
Given: two balls of equal mass with the velocities shown collide, coefficient of rest = 0.8
Find: velocities after impact
. .
45° 4
35 m/s 10 m/s
x - mom.: u1 + u2 = 5/√2 - 8
Rest: u2 - u1 = 0.8 [(5/√2) + 8 ]
u2 = 2.76 u1 = -7.23 5/√2
5/√2 6
8
5/√2
u1
6
u2
x
y
Work / Energy :
T1 + V1 + WNC = T2 + V2
where T = m vc2 = Ic ω2 = Ic ω2
Vgrav = -mgh
Vspring = k x2 where x = l - lo= stretch
1 2
1 2
1 2
1 2
31
61
v
A
!=15°µ!!=0.3k
B
10 m
Work-Energy Example ref ~Meriam & Kraige 3/11 Given: • A crate of mass m slides down an incline • m =50 kg, θ=15°, µk=0.3, • Reaches A with speed 4 m/s
Find: (A) Speed of crate vB as it reaches a point B
10 m down the incline from A
Solution: • Rectilinear motion, align axes accordingly -
i.e. || & ⊥ to incline • FBD of crate in general position (working
over a motion interval here)
Fy! = mg cos" #N = 0 $ N = mgcos"
• No movement ⊥ to incline so
y
x
The
θ
θ
m g
N F µ
62
Work-Energy Example ref ~Meriam & Kraige 3/11 Solution (cont’d): • Work done is due to the resultant forces in
direction of displacement (i.e. down incline) & includes Friction & component of Weight
• Principle of Work-Energy then says
U A-B = (mg sin! "Nµk )#xAB
= (mg sin! "mg cos!µk )#xAB
UA-B = !TA-B = TB " TA ! TB = UA -B + TA12mvB
2 = mg(sin! " cos!µk )#xAB +12mvA
2
vB = 2g(sin! " cos!µk )#xAB + vA2
!
vB = 2*9.81(m / s2 )*(sin15°" cos15°*0.3)*10m+ (4m / s)2
vB = 3.15 m / s
v
A
!=15°µ!!=0.3k
B
10 my
x
The
θ
θ
m g
N F µ
32
63
Work-Energy: Example ref ~Meriam & Kraige 3/13 Given: • Block (m =50 kg) mounted on rollers • Massless spring w/ k=80 N/m • Released from rest at A where spring has
initial stretch of 0.233 m • Cord w/ constant tension P=300 N attaches
to block & routed over frictionless/massless (ideal) pulley @ C
Find: (A) Speed of block vB as it reaches a point B
directly under the pulley.
Solution: • Again, rectilinear motion, align axes
accordingly • FBD of block in general position (working
over a motion interval here)
• Look at alternative - include the rope in as part of the SYSTEM - reduce FDB to an ACTIVE Force Diagram!
k P=300N x A 1.2 m 0.9 m A B C
P
Fs
mg
R
N
P
Fs
mg
N
!
k The image part with relationship ID rId10 was not found in the file.
P=300N
x A
1.2 m
0.9 m
A B
C lA
ps
y
x
64
Work-Energy: Example ref ~Meriam & Kraige 3/13 Solution (cont’d):
k P=300N x A 1.2 m 0.9 m A B C
y
x
P
FsActive Force
Diagram
ACTIVE Force Diagram!
• Eliminate Normal Forces ⊥ to displacement @ their point of contact {THEY DO NO WORK!}
P
Fs
mg
R
N– Weight (mg) & Roller reactions (N) – Pulley force on rope (R)
• Active forces DO work on the system – Spring Force (Fs) => opposes motion
Fs = !kx
– Assuming block can actually reach B
UABFs = FsdxxA
xB
! = "kxdxxA
xB
! =12kx 2
xA
xB
= " 12k(xB
2 " xA2 )
UABFs = !1280(N / m){(1.2 + 0.233)2 ! 0.2332}(m 2) = !80Joules
k The image part with relationship ID rId16 was not found in the file.
P=300N
x A
1.2 m
0.9 m
A B
C lA
ps
33
65
Work-Energy: Example ref ~Meriam & Kraige 3/13 Solution (cont’d):
k P=300N x A 1.2 m 0.9 m A B C
y
x
P
FsActive Force
Diagram
P
Fs
mg
R
N
Lcord = sP + l = constant!sP = "!l = lA " lB
= 1.22 + 0.92 " 0.9 # 0.61m
– Cord Tension (P) => constant – Displacement of P
• Calculate Work done on system
UABP= P!s = 300(n) * 0.61(m)
=180Joules
lA
= lB
• Work-Energy UTOT = !TA-B = TB " TA
!80 +180(Joules) = 12m(vB
2 ! 0)
! vB =100(Joules) * 2
50Kg= 2.0m / s
ps
k The image part with relationship ID rId24 was not found in the file.
P=300N
x A
1.2 m
0.9 m
A B
C lA
ps
66
Conservation-Energy Example ref ~Meriam & Kraige 3/17 Given: • m =3 kg slider on circular track shown • Starting from A with vA=0 • lo=0.6 m (unstretched), k=350 N/m • µ=0 (i.e friction is negligible)
Find: (A) Velocity of slider as it passes B
Solution: • FBD of crate in general position (working
over a motion interval here)
• Identify – Conservative Forces
NFs
mgFBD
mg (Weight/Gravity) & Fs (Spring) – Non-working Constraint Forces
N (Track reaction force)
The image part with relationship ID rId7 was not found in the file.The image part with relationship ID rId8 was not found in the file.The image part with relationship ID rId9 was not found in the file.The image part with relationship ID rId10 was not found in the file.
A R=0.6m B v B 0.6m k =350 N/m
AR=0.6m
B vB
0.6m
k=350 N/m
y
34
67
A R=0.6m B v B 0.6m k =350 N/m
Conservation-Energy Example ref ~Meriam & Kraige 3/17 Solution: • ALL Forces are either Conservative or
Non-working constraints, therefore Cons. Of Energy applies!
• Pulling together all components & isolating vB NFs
mgFBD
!ETOT = !T + !Vg + !Ve = 0
!TAB =12m(vB
2 " vA2 ) = 1
2m(vB
2 " 0)
!VABg = mg(yB " yA ) = mg(0 " R)
!VABe =12k (lB " l0 )
2 " (lA " l0 )2{ }
vB = 2gR + km
R2 ! ( 2R ! R)2{ }• Incorporating numerical values of all terms
!
vB = 2 * 9.81(m /s2)(0.6m) +350N /m
3kg(0.6m)2 " ( 2 *0.6m " 0.6m)2{ } = 6.82 m /s
AR=0.6m
B vB
0.6m
k=350 N/m
y
Work/ Energy T1 + V1 + WNC = T2 = V2
Given: k, m, Θ block released from rest with spring compressed δ
T1 = 0 V1= 1/2 k δ2
WNC = -Fd
where N = mg cosΘ, F = µkN while sliding
T2=0 V2 = 1/2k(d - δ)2 - mgd sin Θ
Find: distance travelled to stop, d
k
m
Θ
35
Given: k, m, R, Θ, I released from rest, no slip, no initial deflection in spring
Find: speed after 1/2 revolution
Θ
m,R
T1 = 0 V1 = 0 WNC = 0
T2 = 1/2 m v22 + 1/2 I ω2
2, V2 = 1/2 k (π R)2 - mg π R sin Θ need also v2 = R ω2
(no slip condition) 0 = 1/2 m v2
2 + 1/2 I (v2/ R)2 + 1/2 k (π R)2 - mg π R sin Θ
πR k
Kinetics Summary • Three general solution approaches for establishing the governing
equations of motion (EOM) => Which one to use? i) Newton’s Laws ii) Work- Energy & Conservation of Energy iii) Impulse - Momentum & Conservation of Momentum
MP! = ICG " + reff maCG
UA!B = matdssA
s B
" = mvdvvA
vB
" = 12m vB
2 ! vA2( ) = #TA!B
– Typical forces • Springs • Friction • Gravitation
F f = µs / k NF = k s - s0( )
F =mg
UNC = !T + !Vg + !Ve = !ETOT
I = FRdt! = dL! = "L
F! = m aCG
36
Given: Box placed on conveyor with zero initial velocity.
µk
VBelt
Find: Time during which slip occurs
Find: Time during which slip occurs
VB= 10 ft/s
µ = 0.333
ΣF↑ = N - mg = 0
ΣF→ = µN = ma µmg = ma a= µg
v = µgt
slip stops where v = vB
so t = vB/ µg
mg ma
N F = µN
during slip
once slip stops F = 0 VBox
VB
t
37
Given: ball of radius r released from rest on incline, no slip, Θ = 30°
Find: acceleration
r
Θ
ΣF = N - mg cos 30 = 0
ΣF = mg sin 30 - F = ma
ΣMG = rF = 2/5 mr2α
→
if no slip, a = rα so mg sin 30 - 2/5 mrα = mrα
so α = (5/7)(g/r)(1/2); a = (5g/7)(1/2) = 5g/14
v = a t = 5gt 14
also: N = mg → Fmax = µsN = √3 2
√3 µ mg 2
F = mrα = 2 mg 5 7
F < Fmax requires < 1 7
√3 µ 2
or µ > 2 7 √3
N
F
30°
mg G c G
ma c
Iα
38
Given: given v , r
Find: Φ; tension
ma = m r ω2
ma = m v2/r
rv
Φ
Φ
ma
mg
T
ΣF ↑ = T cos Φ - mg = 0 ΣF → = T sin Φ = m v2/r
or ΣF = mg sin Φ = ma cos Φ
so tan Φ = = ag
v2 gr
Given: m1 released from rest, strikes m2 Find: max spring compression m1
m2 k
State 1 as shown State 2
vp
d
39
State 3
State 4
vʹ′p vm
δ
Vibrations
x
k c m
kx
cx .
mx ..
Σ F = - kx - cx = mx . ..
Or mx + cx + kx = 0 .. .
x + cx / m + kx / m = 0 .. .
general x + 2 ζ ω x + ω2 x = 0 ..
where w = natural freq. ζ = damping factor
ζ < 1 underdamped
ζ > 1 overdamped
ζ = 1 critically damped
40
!F
Solution:!• Rectilinear motion! Constrained vertically so
align axes accordingly!
Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!Given:!• A collar of mass m slides vertically on a shaft with
kinetic coefficient of friction, µk.!• Applied force F is constant but its direction varies as!! ! !θ = k t k=> constant
• Collar starts from rest @ θ=0°!Find:!(A)! Magnitude of F which results in collar
coming to rest at θ=90°. !
• FBD of collar in general position (working over a motion interval here)!
Fx! = Fsin" # N = 0 !
Fy! = #F cos" +mg +Nµk =may
• Newtonʼs Laws!! N = Fsin"
& Fm
(sin"µk # cos") + g = ay
x
y
θ!F!
µkN!
mg!
N
!F
θ!F!
µkN!
mg!
x
y
N
Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!
!
dvy0
vyf" = {Fm(sin(kt)µk # cos(kt)) + g}dt
0
t"
!
vyf =Fmk
("cos(kt)µk " sin(kt)) + gt# $ %
& ' ( 0
t
• Donʼt know F but DO know itʼs constant!!• Angle-time relation (θ = k t) cleans up!
– Kinematic relationship variables & !– Proper (Pos,Vel )BCʼs for integration!
• Starts: vyi=0, θ=0 => t=0!• Ends: vyf=0, θ=π/2 => t=π/(2k)!
– Start w/ general upper limits!
!
Fm(sin"µk # cos") + g = ay =
dvydt
Solution (Contʼd):
!
vyf =Fmk
{[1" cos(kt)]µk " sin(kt)}+ gt
41
!F
θ!F!
µkN!
mg!
x
y
N
Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!
!F =" mg
2(1 # µk )
Solution (continued):!• Now using the final BC for t=π/2k, vyf=0
Langiappe:!• What are:!
– The collarʼs vertical displacement as a function of time?!– The collarʼs total distance traveled?
!
vyf =Fmk
{[1" cos(kt)]µk " sin(kt)}+ gt
!
0 =Fmk{[1" cos(# /2)]µk " sin(# /2)}+
g#2k
!
0 =Fmk{µk "1}+
g#2k
!F
θ!F!
µkN!
mg!
x
y
N
Particle Kinetics: Cartesian Example ref ~Meriam & Kraige 3/5!
Langiappe (continued):!• Returning to the expression for vy=f(t)
!
vyf =dydt
=Fmk
{[1" cos(kt)]µk " sin(kt)}+ gt
!
dy0
y f" =Fmk{[1# cos(kt)]µk # sin(kt)}+ gt
$ % &
' ( ) dt
0
t"
!
y f =Fmk
{[t " sin(kt)]µk + cos(kt)}+12gt 2
#
$ % &
' ( 0
t
!
y f =Fmk{[t " sin(kt)]µk + cos(kt) -1}+
12gt 2
!
y f =Fmk{[ "2k
# sin("2)]µk + cos("
2) -1}+
12g "2$
% &
'
( ) 2
!
y f =Fmk{[ "2k
#1]µk #1}+12g "2$
% &
'
( ) 2
42
!
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/7!
Given:!• A box of mass m (particle) is released
from rest @ top of a smooth circular track.!
Find:!(A) !Normal force N as a function of
position θ along the circular track.!!
Solution:!• FBD of box in general position - working over
motion interval A-B, then instantaneous @ B!!– Track smooth ==> µ=0, no friction!!
Fn! = N "mgsin# = manFt! = mgcos# = mat
• Attach normal-tangential coordinate axes!
• Newtonʼs Laws for general θ!
A
R
r
B
!
(B) !The angular velocity (ω) of the pulley B such that the boxes donʼt slide onto the conveyor belt !
N
n
t
mg
!
N!
ω!
• Must use kinematics to resolve unknowns!!
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/7!
Solution (contʼd):!• Kinematics: Circular track-> path coord!
Fn! = N "mgsin# = man = mv2
R
Ft! = mgcos# = mat = m v•
= m s••
• Using tangential direction to resolve velocity by integrating alternate form => vdv=atds!
• Now using the normal direction!
A
R
r
B
!
N
n
t
mg
!at = s••= gcos! s = R! " ds = Rd!
12v2 = gRsin! 0
!= gRsin!
vdv0
v
! = gRcos "d"0
"
!
v2 = 2gRsin!
! N = m v2
R+mgsin"
= 2mgsin" +mgsin"
(A) N = 3mgsin!
N!
ω!
s = R θ!
43
Particle Kinetics: Path Coord Example ref ~Meriam & Kraige 3/7!
Solution (contʼd):!• Belt Kinematics:!
– From previous page, the box speed as it reaches B!
– Belt must move at the same speed to avoid slip, i.e. relative velocity between belt/box is zero!
A
R
r
B
!
N
!
s=R!
v = 2gRsin!!=
"2= 2gR
(B) ! =2Rgr
r!
B Bv
vbelt
vB = vbelt2gR =! r
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